Question

A box of mass ? is pushed a distance Δ? across a level floor by a constant applied force ?. The coefficient of kinetic friction between the box and the floor is ?. Assuming the box starts from rest, express the final velocity ?f of the box in terms of ?, Δ?, ?, ?, and ?. A box labeled M sits on the left side of a table. An arrow touching the left side of the box representing the applied force is labeled F and points horizontally rightward. Another box representing the final position of the original box sits on the right side of the table. The distance between the box on the right and the box on the left is labeled delta X.

?f=

Answer 1

Answer 2

The final velocity (${�}_{�}$) of the box can be expressed in terms of the applied force ($�$), the mass of the box ($�$), the coefficient of kinetic friction (${�}_{�}$), the distance it is pushed ($\mathrm{\Delta }�$), and other relevant variables.

The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. In this case, the work done by the applied force will be used to overcome friction and change the kinetic energy of the box.

The work done by the applied force ($�$) is given by:

$�=�\cdot \mathrm{\Delta }�$

The work done against friction is:

${�}_{�}={�}_{�}\cdot �\cdot \mathrm{\Delta }�$

where $�$ is the normal force, which is equal to the weight of the box ($�\cdot �$).

The change in kinetic energy is:

$\mathrm{\Delta }��=\frac{1}{2}�{�}_{�}^2-\frac{1}{2}�\cdot 0^2=\frac{1}{2}�{�}_{�}^2$

Since the work done on the box results in the change in kinetic energy, we can equate the two expressions for work:

$�\cdot \mathrm{\Delta }�-{�}_{�}\cdot �\cdot �\cdot \mathrm{\Delta }�=\frac{1}{2}�{�}_{�}^2$

Now, we can solve for ${�}_{�}$:

${�}_{�}=\sqrt{\frac{2}{�}\cdot (�-{�}_{�}\cdot �\cdot �)\cdot \mathrm{\Delta }�}$

So, the final velocity (${�}_{�}$) of the box, assuming it starts from rest, can be expressed in terms of the applied force ($�$), the mass of the box ($�$), the coefficient of kinetic friction (${�}_{�}$), and the distance it is pushed ($\mathrm{\Delta }�$) as shown above.

Answer 3

The final velocity (${�}_{�}$) of the box can be expressed in terms of the applied force ($�$), the mass of the box ($�$), the coefficient of kinetic friction (${�}_{�}$), the distance it is pushed ($\mathrm{\Delta }�$), and other relevant variables.

The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. In this case, the work done by the applied force will be used to overcome friction and change the kinetic energy of the box.

The work done by the applied force ($�$) is given by:

$�=�\cdot \mathrm{\Delta }�$

The work done against friction is:

${�}_{�}={�}_{�}\cdot �\cdot \mathrm{\Delta }�$

where $�$ is the normal force, which is equal to the weight of the box ($�\cdot �$).

The change in kinetic energy is:

$\mathrm{\Delta }��=\frac{1}{2}�{�}_{�}^2-\frac{1}{2}�\cdot 0^2=\frac{1}{2}�{�}_{�}^2$

Since the work done on the box results in the change in kinetic energy, we can equate the two expressions for work:

$�\cdot \mathrm{\Delta }�-{�}_{�}\cdot �\cdot �\cdot \mathrm{\Delta }�=\frac{1}{2}�{�}_{�}^2$

Now, we can solve for ${�}_{�}$:

${�}_{�}=\sqrt{\frac{2}{�}\cdot (�-{�}_{�}\cdot �\cdot �)\cdot \mathrm{\Delta }�}$

So, the final velocity (${�}_{�}$) of the box, assuming it starts from rest, can be expressed in terms of the applied force ($�$), the mass of the box ($�$), the coefficient of kinetic friction (${�}_{�}$), and the distance it is pushed ($\mathrm{\Delta }�$) as shown above.